2 not unique apparently

• irreducible Z[G] modules,
• Z-forms of irreducible Q[G] modules,
• indecomposable Z[G] modules, or
• indecomposable Z[G] modules that are finitely generated and free as Z-modules.

I'll assume the last is the main concern.

The irreducible modules of ZS3 are all finite and have an elementary abelian p-group as their additive group. For p=2,3 there are 2 each, and for p>3, there are 3 each.

The irreducible CS3 modules are all realizable over Q. Every such module may be realized over Z, but for general finite G and possibly reducible the two-dimensional representation has two distinct Z-forms, giving four total "irreducible" Z-free ZS3 modules, a QG module may have multiple inequivalent that is, four total Z-forms . I think for of irreducible QS3 modulesover S3, they are all unique.

Indecomposable ZS3 modules up to isomorphism are more complicated than the human mind can possibly comprehend. Indeed, even those in which S_3 acts as the identity are much too complex. Luckily they divide up into several types: annihilated by a prime p (then classified by modular representation theory), torsion (more complicated, but basically now p-adic integral reps), Gorenstein projective (Z-free, so covered in the next bullet point), or madness (that is, the rest).

The indecomposable ZS3 modules that are free as Z-modules are classified in:

Lee, Myrna Pike. "Integral representations of dihedral groups of order 2p." Trans. Amer. Math. Soc. 110 (1964) 213–231. MR 156896 doi:10.2307/1993702

There are 10 of them, and the Krull-Schmidt theorem fails for them. Not only are indecomposables not completely reducible, the decomposition of a finitely generated Z-free module into indecomposable summands is not unique. In other words, integral representations of even very small groups are quite complicated.

1

• irreducible Z[G] modules,
• Z-forms of irreducible Q[G] modules,
• indecomposable Z[G] modules, or
• indecomposable Z[G] modules that are finitely generated and free as Z-modules.

I'll assume the last is the main concern.

The irreducible modules of ZS3 are all finite and have an elementary abelian p-group as their additive group. For p=2,3 there are 2 each, and for p>3, there are 3 each.

The irreducible CS3 modules are all realizable over Q. Every such module may be realized over Z, but for general finite G and possibly reducible modules, a QG module may have multiple inequivalent Z-forms. I think for irreducible modules over S3, they are all unique.

Indecomposable ZS3 modules up to isomorphism are more complicated than the human mind can possibly comprehend. Indeed, even those in which S_3 acts as the identity are much too complex. Luckily they divide up into several types: annihilated by a prime p (then classified by modular representation theory), torsion (more complicated, but basically now p-adic integral reps), Gorenstein projective (Z-free, so covered in the next bullet point), or madness (that is, the rest).

The indecomposable ZS3 modules that are free as Z-modules are classified in:

Lee, Myrna Pike. "Integral representations of dihedral groups of order 2p." Trans. Amer. Math. Soc. 110 (1964) 213–231. MR 156896 doi:10.2307/1993702

There are 10 of them, and the Krull-Schmidt theorem fails for them. Not only are indecomposables not completely reducible, the decomposition of a finitely generated Z-free module into indecomposable summands is not unique. In other words, integral representations of even very small groups are quite complicated.